# Using a crankshaft, a carpenter drills a hole ,1 cm in radius, through a wooden ball along a diameter. If the radius of the ball is 4 cm, what is the volume of wood remaining? .

Dec 1, 2017

Calculate the removed portion and subtract it from the original volume.
$256.3 c {m}^{3}$

#### Explanation:

The removed portion is a cylinder with two spherical portion caps. A quick approximation could be done by ignoring the curvature. In that case the cylinder volume is just:
${V}_{c} = \pi \times {r}^{2} \times h$ ; ${V}_{c} = \pi \times 1 \times 4 = 12.6 c {m}^{3}$

The Sphere volume is ${V}_{s} = \frac{4}{3} \pi \times {r}^{3}$
${V}_{s} = \frac{4}{3} \pi \times {4}^{3} = 268.1 c {m}^{3}$

The remaining volume (approximated) is thus
$268.1 c {m}^{3} - 12.6 c {m}^{3} = 255.5 c {m}^{3}$

To find the exact volume removed, add the volume of the caps:
V_cp = pi xx h^2 xx (r – h/3)
h = 0.127 (calculated from chord "a" - cut radius of 1 - see graphic)
V_cp = pi xx 0.127^2 xx (4 – 0.127/3)
${V}_{c} p = 0.2005$
We have to take TWO of those for a total volume of 0.401
The revised cylinder length is 4 - 0.254 = 3.746
${V}_{c} = \pi \times {r}^{2} \times h$ ; ${V}_{c} = \pi \times 1 \times 3.746 = 11.8 c {m}^{3}$
Add this to the caps: $11.8 + 0.401 = 12.17 c {m}^{3}$ (only 3.4% error from the approximation)

The remaining volume (exact) is thus
$268.1 c {m}^{3} - 11.8 c {m}^{3} = 256.3 c {m}^{3}$ (only 0.31% error from the approximation!).

http://mathworld.wolfram.com/SphericalCap.html